SQL Server: select has too many columns for insert - sql-server-2000

Question

I'm trying to run this query and I keep getting this error:
The select list for the INSERT statement contains more items than the insert list. The number of SELECT values must match the number of INSERT columns.
This error comes after I try doing INSERT INTO #tempTable SELECT.... It worked just fine when I had only 2 columns for each temp table, but now that I've added a third, it keeps giving me this error even though it appears to be selecting 3 columns to insert into the table with 3 columns.
The query overall is trying to get a few column values, (customer ids, contacts, and a salesman id) over different tables, but the problems have come from needing to root out duplicate customer IDs and making sure that only 1 salesman and 1 contact showing up for each customer. If it wasn't for the salemen id, the query would work perfectly. Here is what I have so far:
if object_id('tempdb..#tempTable') IS NOT NULL DROP TABLE #tempTable
if object_id('tempdb..#tempTable2') IS NOT NULL DROP TABLE #tempTable2
CREATE TABLE #tempTable(
CustomerID int,
ContactName nvarchar(50),
SalesmenID nvarchar(4)
)
CREATE TABLE #tempTable2(
CustomerID int,
ContactName nvarchar(50),
SalesmenID nvarchar(4)
)
INSERT INTO #tempTable
(CustomerID, ContactName,SalesmenID)
SELECT Customers.[Customer ID],Salesmen.[4 Letter ID],
CASE([Customer Contact].defaultprintonorder)
WHEN 0
THEN 'zzzzzzz_NOCONTACT'
ELSE
[Customer Contact].[Contact Name]
END as ContactName
From Customers
LEFT JOIN [Customer Contact] on Customers.[Customer ID]=[Customer Contact].[Customer ID]
Left Join [Customer Salesmen] On Customers.[Customer ID]=[Customer Salesmen].[Customer ID]
INNER JOIN Salesmen on [Customer Salesmen].[Salesman Name]=Salesmen.[Salesman Name]
WHERE Customers.[Customer ID] NOT IN(SELECT CustomerID FROM #tempTable)
GROUP BY Customers.[Customer ID], [Contact Name], DefaultPrintOnOrder
INSERT INTO #tempTable2 (CustomerID, ContactName,SalesmenID)
SELECT distinct CustomerID, '', SalesmenID FROM #tempTable
UPDATE #TempTable2 SET
#tempTable2.CustomerID=#tempTable.CustomerID,
#tempTable2.ContactName=#tempTable.ContactName
FROM
#TempTable2
INNER JOIN #TempTable ON #TempTable2.CustomerID=#TempTable.CustomerID
SELECT Salesmen.[4 Letter ID],
[Customers].[Customer ID],
[Customer Contact].[Contact Name]
FROM Customers
Right JOIN #TempTable2 ON
Customers.[Customer ID]=#TempTable2.CustomerID
Right JOIN [Customer Salesmen] ON
#TempTable2.CustomerID=[Customer Salesmen].[Customer ID]
INNER JOIN
[Salesmen] ON
[Customer Salesmen].[Salesman Name]=Salesmen.[Salesman Name]
LEFT JOIN
[Customer Contact] ON
#TempTable2.[CustomerID]=[Customer Contact].[Customer ID]
EDIT:
I added SalesmenID to the inserts, but now I'm getting this error message 3 times:
Invalid column name 'SalesmenID'.
It comes up once for the temptable insert and twice for the temptable2 insert

Answer 1

You're trying to store 3 values (Customers.[Customer ID], Salesmen.[4 Letter ID], ContactName) in 2 fields (CustomerID, ContactName)
INSERT INTO #tempTable
(CustomerID, ContactName)
SELECT Customers.[Customer ID],Salesmen.[4 Letter ID],
CASE([Customer Contact].defaultprintonorder)
WHEN 0
THEN 'zzzzzzz_NOCONTACT'
ELSE
[Customer Contact].[Contact Name]
END as ContactName

Answer 2

Are you sure the problem is where you indicated?
This insert has 2 destination columns and 3 select columns:
INSERT INTO #tempTable
(CustomerID, ContactName) -- 2 columns
SELECT Customers.[Customer ID], --column 1
Salesmen.[4 Letter ID], --column 2
CASE([Customer Contact].defaultprintonorder) -- column 3!
WHEN 0
THEN 'zzzzzzz_NOCONTACT'
ELSE
[Customer Contact].[Contact Name]
END as ContactName
From Customers...